Solve (5 questions)?

(Only use partial fractions where possible)

  1. \intx\sinx\cosxdx
  2. \int(x^P)/(x(x^(2P)+1)dx
  3. \int(\lnx)/(x\sqrt(1+(\lnx)^2)dx
  4. \int\sqrt(1+e^x)dx
  5. \int(x+\sin^-1(x))/(\sqrt(1-x^2))dx

2 Answers
Andrea S.
Mar 7, 2018

int xsinxcosxdx = (sin2x-2xcos2x)/8 +C

int x^P/(x(x^(2P)+1)) dx = 1/P arctanx^P +C

int lnx/(xsqrt(1+(lnx)^2))dx = sqrt(1+(lnx)^2)+C

int sqrt(1+e^x)dx = 2sqrt(1+e^x) +ln ((sqrt(1+e^x)-1)/(sqrt(1+e^x)+1))+C

int (x+arcsinx)/sqrt(1-x^2)dx= 1/2arcsin^2x - sqrt(1-x^2)+C

Explanation:

Not sure we really need partial fractions except in one case:

(1)

int xsinxcosxdx = 1/2 int xsin2x dx

integrate by parts:

int xsinxcosxdx = -1/4 int xd (cos2x)

int xsinxcosxdx = -(xcos2x)/4 + 1/4 int cos2xdx

int xsinxcosxdx = -(xcos2x)/4 + 1/8 sin2x +C

int xsinxcosxdx = (sin2x-2xcos2x)/8 +C

(2)

int x^P/(x(x^(2P)+1)) dx = int x^(P-1)/(x^(2P)+1) dx

Substitute t=x^P, dt = Px^(P-1)

int x^P/(x(x^(2P)+1)) dx = 1/P int dt/(t^2+1) = 1/P arctant+C

int x^P/(x(x^(2P)+1)) dx = 1/P arctanx^P +C

(3)

int lnx/(xsqrt(1+(lnx)^2))dx

Substitute t = 1+(lnx)^2, dt =2lnx/x dx

int lnx/(xsqrt(1+(lnx)^2))dx = int dt/(2sqrt(t)) =sqrtt + C

int lnx/(xsqrt(1+(lnx)^2))dx = sqrt(1+(lnx)^2)+C

(4)

int sqrt(1+e^x)dx

Substitute t=sqrt(1+e^x), dt= e^x/(2sqrt(1+e^x))dx =(t^2-1)/(2t)dx

int sqrt(1+e^x)dx = 2int t^2/(t^2-1)dt

int sqrt(1+e^x)dx = 2int (t^2-1+1)/(t^2-1)dt

int sqrt(1+e^x)dx = 2int dt + 2int dt/(t^2-1)

int sqrt(1+e^x)dx = 2t + 2int dt/((t-1)(t+1))

2/((t-1)(t+1)) = A/(t-1)+B/(t+1)

2= A(t+1)+B(t-1) = (A+B)t+(A-B)

{(A+B = 0),(A-B=2):}

{(A=1),(B=-1):}

int sqrt(1+e^x)dx = 2t + int dt/(t-1)- int dt/(t+1)

int sqrt(1+e^x)dx = 2t + ln abs (t-1)- ln abs (t+1) +C

int sqrt(1+e^x)dx = 2sqrt(1+e^x) +ln ((sqrt(1+e^x)-1)/(sqrt(1+e^x)+1))+C

(5)

int (x+arcsinx)/sqrt(1-x^2)dx

int x/sqrt(1-x^2)dx+int arcsinx/sqrt(1-x^2)dx

int x/sqrt(1-x^2)dx = - int (d(1-x^2))/(2sqrt(1-x^2)) = -sqrt(1-x^2)+C

int arcsinx/sqrt(1-x^2)dx = int arcsinx d(arcsinx) = 1/2arcsin^2x+C

int (x+arcsinx)/sqrt(1-x^2)dx= 1/2arcsin^2x - sqrt(1-x^2)+C

Noah G
Mar 7, 2018

int xsinxcosx dx = 1/8sin(2x) - 1/4xcos(2x) + C

Explanation:

Here is the answer to 1. I would rewrite and then use integration by parts. Recall that sin(2x) = 2sinxcosx.

I = int x(1/2sin(2x))dx

I = int (xsin(2x))/2dx

I = 1/2int xsin(2x) dx

Now let u= x and dv = sin(2x)dx. We can immediately see that du = dx. Integrating dv is a little harder. Letting n = 2x, we get dn = 2dx and dx= (dn)/2. Therefore v = -1/2cosn = -1/2cos(2x)

Now recall that integration by parts is

int udv =uv - int vdu

int xsin(2x)dx = -1/2xcos(2x) - int -1/2cos(2x)dx

int xsin(2x)dx= -1/2xcos(2x) + int 1/2cos(2x)dx

int xsin(2x)dx= -1/2xcos(2x) + 1/4sin(2x) + C

1/2int xsin(2x)dx= 1/8sin(2x) - 1/4xcos(2x) + C

Hopefully this helps!

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